How do I to find the limits of these? $$\lim _{x\to \frac{\pi }{4}}\left(\frac{1-\sin 2x}{\cos x -\sin x}\right)$$
$$\lim _{x\to \frac{\pi}{4}}\left(\frac{\cos^5x - \sin^5x}{\cos\:2x}\right)$$
I tried to solve but I get 0/0.
Help me to find the limit.
 A: Very easy. Just factor out. 
$$1-2\sin{x}=\sin^2x+\cos^2x-2\sin{x}=\sin^2x+\cos^2x-2\sin{x}\cos{x}=(\cos{x}-\sin{x})^2$$
Now just divide and get $\quad\cos{x}-\sin{x}\quad$ which will equal $\quad0$.
The other problem is the same. Can you figure it out on your own? A little hint:
$$\cos^5x-\sin^5x=(\cos{x}-\sin{x})(\cos^4x+\cos^3{x}\sin{x}+\cos^2x\,\sin^2{x}+\cos{x}\sin^3{x}+\sin^4{x})$$
A: For the first limit write the numerator as $\cos^2{x}+\sin^2{x}-2\sin{x}\cos{x}=\left(\cos{x}-\sin{x}\right)^2$ where we have used $\sin{2x}=2\sin{x}\cos{x}$ and $\cos^2{x}+\sin^2{x}$. The fraction simplify to
$$\cos{x}-\sin{x}\to 0$$
when $x\to\pi/4$
For the second limit we use
$$\cos{2x}=\cos^2{x}-\sin^2{x}=(\cos{x}-\sin{x})(\cos{x}+\sin{x})$$
and
$$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$
to simplify the fraction to
$${\cos^4{x}+\cos^3{x}\sin{x}+\cos^2{x}\sin^2{x}+\cos{x}\sin^3{x}+\sin^4{x}\over \cos{x}+\sin{x}}$$
and the limit of this fraction is
$${5\sqrt{2}\over 8}$$
A: $$
\begin{aligned}
\lim _{x\to \frac{\pi }{4}}\left(\frac{1-\sin2x}{\cos x\:-\sin x}\right)
& = \lim _{t\to 0}\left(\frac{1-sin2\left(t+\frac{\pi }{4}\right)}{\cos \left(t+\frac{\pi \:}{4}\right)\:-\sin \left(t+\frac{\pi \:}{4}\right)}\right)
\\& = \lim _{t\to 0}\left(-\frac{\sqrt{2}\left(1-\sin \left(2\right)\left(t+\frac{\pi \:}{4}\right)\right)}{2\sin \left(t\right)}\right)
\\& = \color{red}{\infty}
\end{aligned}
$$
$$
\begin{aligned}
\lim _{x\to \frac{\pi }{4}}\left(\frac{\cos ^5x\:-\:\sin ^5x}{\cos \:2x}\right)
& = \lim _{t\to 0}\left(\frac{\cos ^5\left(t+\frac{\pi }{4}\right)\:-\:\sin ^5\left(t+\frac{\pi \:}{4}\right)}{\cos \left(2\left(t+\frac{\pi \:}{4}\right)\right)}\right)
\\& = \lim _{t\to 0}\left(\frac{-\frac{\sqrt{2}}{4}\sin ^5\left(t\right)-\frac{\sqrt{2}\cdot 5}{2}\sin ^3\left(t\right)\cos ^2\left(t\right)-\frac{\sqrt{2}\cdot 5}{4}\cos ^4\left(t\right)\sin \left(t\right)}{\cos \left(2\left(t+\frac{\pi \:}{4}\right)\right)}\right)
\\& \approx \lim _{t\to 0}\left(\frac{-\frac{\sqrt{2}}{4}\left(t\right)^5-\frac{\sqrt{2}\cdot \:\:5}{2}\left(t\right)^3\left(1-\frac{t^2}{2}\right)^3-\frac{\sqrt{2}\cdot \:\:5}{4}\left(1-\frac{t^2}{2}\right)^4\left(t\right)}{\cos \:\left(2\left(t+\frac{\pi \:}{4}\right)\right)}\right)
\\& = \lim _{t\to 0}\left(-\frac{\sqrt{2}\left(15t^9-80t^7+104t^5-80t\right)}{64\sin \:\left(2t\right)}\right)
\\& \approx \lim _{t\to 0}\left(-\frac{\sqrt{2}\left(15t^9-80t^7+104t^5-80t\right)}{64\left(2t\right)}\right)
\\& = \color{red}{\frac{5}{4\sqrt{2}}}
\end{aligned}
$$
